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Let $f:\Bbb R \to \Bbb R$ be differentiable on $\Bbb R$ and $|f'(x)| \leq \frac {1}2$ for all $x\in \Bbb R$. Let $a \in \Bbb R$ and define $x_1 = a, x_{n+1} = f(x_n)$ for all $n \in \Bbb N$. Show that $\{x_n\}$ is a Cauchy sequence. And show that $\exists x \in \Bbb R$ such that $f(x) = x$.

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  • $\begingroup$ Use the mean value theorem to show that you can apply the Banach fixed point theorem. $\endgroup$
    – Hello
    Nov 24, 2016 at 12:37

2 Answers 2

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By the mean value theorem, for every $x,y\in \Bbb R$, there exists $z\in[x,y]$ such that $$\frac{|f(x)-f(y)|}{|x-y|}=|f'(z)|\leq\frac{1}{2}.$$ It follows that $|f(x)-f(y)|\leq \frac{1}{2}|x-y|$ for every $x,y\in \Bbb R$. Thus, $f$ is a strict contraction and by the Banach contraction principle, there exists a unique $x\in \Bbb R$ such that for every $x_0=a\in \Bbb R$, the sequence $x_{n+1}=f(x_n)$, $n\in \Bbb N$ converge to $x$.

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$f$ is differentiable on $\mathbb R.$ so, we can use MVT at any interval.

$\forall n\in \mathbb N$

$$|x_{n+1}-x_n|=|f(x_n)-f(x_{n-1})|=$$

$$|x_n-x_{n-1}|f'(c|\leq \frac{1}{2}|x_n-x_{n-1}|$$

.

.

$$\leq \frac{1}{2^n}|x_1-x_0|.$$

From here, by triangular inequality, we have

$\forall \; n>p\geq 0$

$$|x_{n+p}-x_n|\leq \frac{ |x_1-x_0|}{2^n}\sum_{k=0}^{p-1}\frac{1}{2^k}$$

$$=2\frac{|x_1-x_0|}{2^n}(1-\frac{1}{2^p})\leq \frac{|x_1-x_0|}{2^{n-1}}.$$

Now, you conclude.

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  • $\begingroup$ There seems to be a typo: I think $|x_n-x_{n-1}|f'(c|$ should be $|x_n-x_{n-1}| |f'(c)|$ $\endgroup$
    – Hello
    Nov 24, 2016 at 16:59

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